Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`
Advertisements
उत्तर
We know `lim_(x -> 0) (sin x)/x` = 1
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = lim_(alpha -> 0) (sin (alpha^"n"))/(1/alpha^"n" (alpha^"n")) xx (alpha^(1/"m") * alpha^"m")/(sin alpha)^"m"`
= `lim_(alpha -> 0) alpha^"n" * (sin(alpha^"n"))/alpha^"n" xx 1/alpha^"m" * 1/(((sinalpha)^"m")/(alpha^"m"))`
= `lim_(alpha -> 0) alpha^("n" - "m") *(sin(alpha^"n"))/alpha^"n" xx 1/((sinalpha)/alpha)^"m"`
= `(lim_(alpha -> 0) alpha^("n" - "m")) xx (lim_(alpha^"n" -> 0) (sin(alpha^"n"))/alpha^"n") xx 1/(lim_(alpha -> 0) ((sin alpha)/alpha)^"m")`
Case (i) m = n
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("m" - "m")) (lim_(alpha^"m" -> 0) (sin(alpha^"m"))/alpha^"m") xx 1/(lim_(alpha -> 0) ((sin alpha)/alpha)^"m")`
= `(lim_(alpha -> 0) alpha^0) xx 1 xx 1/1`
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = 1 xx 1 xx 1` = 1
Case (ii) m > n then n – m < 0
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("n" - "m")) (lim_(apha^"m" -> 0) (sin(alpha^"m"))/alpha^"m") xx 1/((lim_(alpha-> 0) ((sin alpha)/alpha)^"m"))`
= `lim_(alpha -> 0) 1/(alpha^("m" - "n")) xx 1 xx 1/1`
= `oo xx 1 xx 1 = oo`
Since `lim_(alpha -> 0) 1/(alpha^("m" - "n")) = lim_(alpha -> 0) (1/0)^("m" - "n") = oo`
Case (iii) m < n then n – m > 0
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("n" - "m")) (lim_(alpha^"m" -> 0) (sin(alpha^"m"))/(alpha^"m")) xx 1/(lim_(alpha-> 0) ((sin alpha)/alpha)^"m"`
= `(0)^("n" - "m") xx 1 xx 1`
= 0
∴ `lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = {{:(1, "if", "m" = "n"),(oo, "if", "m" > "n"),(0, "if", m < n):}`
APPEARS IN
संबंधित प्रश्न
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 1) (x^2 + x + 1)` = 3
Evaluate the following :
Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`
Evaluate the following :
`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 2) (x - 2)/(x^2 - 4)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.25641 | 0.25062 | 0.250062 | 0.24993 | 0.24937 | 0.24390 |
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) sin x/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.99833 | 0.99998 | 0.99999 | 0.99999 | 0.99998 | 0.99833 |
Evaluate the following limits:
`lim_(x ->) (x^"m" - 1)/(x^"n" - 1)`, m and n are integers
Evaluate the following limits:
`lim_(sqrt(x) -> 3) (x^2 - 81)/(sqrt(x) - 3)`
Evaluate the following limits:
`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + x) - 1)/x`
Evaluate the following limits:
`lim_(x -> oo)(1 + 1/x)^(7x)`
Evaluate the following limits:
`lim_(x -> oo)(1 + "k"/x)^("m"/x)`
Evaluate the following limits:
`lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/x`
Evaluate the following limits:
`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`
Choose the correct alternative:
If `f(x) = x(- 1)^([1/x])`, x ≤ 0, then the value of `lim_(x -> 0) f(x)` is equal to
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
Choose the correct alternative:
`lim_(x -> oo) (1/"n"^2 + 2/"n"^2 + 3/"n"^2 + ... + "n"/"n"^2)` is
Choose the correct alternative:
`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =
`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.
